Solve -2=left(a^2-8right)1^2+a

$$-2 = \left( { a }^{ 2 } -8 \right) { 1 }^{ 2 } +a$$

Solve for a

$$a=-3$$

$$a=2$$

Steps Using Factoring

Calculate 1 to the power of 2 and get 1.

$$-2=\left(a^{2}-8\right)\times 1+a$$

Use the distributive property to multiply a^{2}-8 by 1.

$$-2=a^{2}-8+a$$

Swap sides so that all variable terms are on the left hand side.

$$a^{2}-8+a=-2$$

Add 2 to both sides.

$$a^{2}-8+a+2=0$$

Add -8 and 2 to get -6.

$$a^{2}-6+a=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$a^{2}+a-6=0$$

To solve the equation, factor a^{2}+a-6 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=1 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.

$$\left(a-2\right)\left(a+3\right)$$

To find equation solutions, solve a-2=0 and a+3=0.

$$a=2 a=-3$$

Steps Using Factoring By Grouping

Calculate 1 to the power of 2 and get 1.

$$-2=\left(a^{2}-8\right)\times 1+a$$

Use the distributive property to multiply a^{2}-8 by 1.

$$-2=a^{2}-8+a$$

Swap sides so that all variable terms are on the left hand side.

$$a^{2}-8+a=-2$$

Add 2 to both sides.

$$a^{2}-8+a+2=0$$

Add -8 and 2 to get -6.

$$a^{2}-6+a=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$a^{2}+a-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite a^{2}+a-6 as \left(a^{2}-2a\right)+\left(3a-6\right).

$$\left(a^{2}-2a\right)+\left(3a-6\right)$$

Factor out a in the first and 3 in the second group.

$$a\left(a-2\right)+3\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(a+3\right)$$

To find equation solutions, solve a-2=0 and a+3=0.

$$a=2 a=-3$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 2 and get 1.

$$-2=\left(a^{2}-8\right)\times 1+a$$

Use the distributive property to multiply a^{2}-8 by 1.

$$-2=a^{2}-8+a$$

Swap sides so that all variable terms are on the left hand side.

$$a^{2}-8+a=-2$$

Add 2 to both sides.

$$a^{2}-8+a+2=0$$

Add -8 and 2 to get -6.

$$a^{2}-6+a=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$a^{2}+a-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-1±\sqrt{1^{2}-4\left(-6\right)}}{2}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$a=\frac{-1±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$a=\frac{-1±\sqrt{25}}{2}$$

Take the square root of 25.

$$a=\frac{-1±5}{2}$$

Now solve the equation a=\frac{-1±5}{2} when ± is plus. Add -1 to 5.

$$a=\frac{4}{2}$$

Divide 4 by 2.

$$a=2$$

Now solve the equation a=\frac{-1±5}{2} when ± is minus. Subtract 5 from -1.

$$a=-\frac{6}{2}$$

Divide -6 by 2.

$$a=-3$$

The equation is now solved.

$$a=2 a=-3$$

Steps for Completing the Square

Calculate 1 to the power of 2 and get 1.

$$-2=\left(a^{2}-8\right)\times 1+a$$

Use the distributive property to multiply a^{2}-8 by 1.

$$-2=a^{2}-8+a$$

Swap sides so that all variable terms are on the left hand side.

$$a^{2}-8+a=-2$$

Add 8 to both sides.

$$a^{2}+a=-2+8$$

Add -2 and 8 to get 6.

$$a^{2}+a=6$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+a+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+a+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$a^{2}+a+\frac{1}{4}=\frac{25}{4}$$

Factor a^{2}+a+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$a+\frac{1}{2}=\frac{5}{2} a+\frac{1}{2}=-\frac{5}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$a=2 a=-3$$